Does this Infinite summation of Bessel function has a closed form?

Question

The summation is $$\sum_{n>0}\frac{J_n^2(x)}{n}\sin\frac{2n\pi}{3}$$ I found a thread Infinite sum of Bessel Functions and a wiki article here may be helpful. However, I still cannot figure this out. Also see the description of Bessel function of first kind at Mathworld, equation 61-66 may be helpful.

Answer 1

Neumann's addition theorem is given by \begin{align} J_{0}\left(\sqrt{x^{2} + y^{2} - 2 x y \cos\phi}\ \right) = J_{0}(x) J_{0}(y) + 2 \sum_{n=1}^{\infty} J_{n}(x) J_{n}(y) \cos(n\phi). \end{align} By letting $y=x$ it can quickly be sen that \begin{align} J_{0}\left(2 x \sin(\phi/2) \right) = J_{0}^{2}(x) + 2 \sum_{n=1}^{\infty} J_{n}^{2}(x) \cos(n\phi). \end{align} Integrate both sides with respect to $\phi$ to obtain the expression \begin{align} \sum_{n=1}^{\infty} \frac{1}{n} \, J_{n}^{2}(x) \sin\left(\frac{2 n\pi}{3}\right) = \frac{\pi}{3} \, J_{0}^{2}(x) + \frac{1}{2}\int_{0}^{2\pi/3} J_{0}\left( 2 x \sin\left(\frac{\phi}{2}\right)\right) \, d\phi. \end{align}